The total distance traveled by an object divided by the total amount of time needed to travel equals the?

Using software, a simulation of a javelin being thrown both on Earth and on the moon is created. In both cases, the velocity and

mario62 [17]

Horizontal distance covered by a projectile is X = Vix *T

where Vix is the initial horizontal component of velocity and T is time taken by the projectile

Vix = ViCos theta

In question they said that initial velocity and angle is same on earth and moon

so Vix would remains same

now let's see about time taken T

time taken to reach the highest point

Vfy = Viy +gt

at highest point vertical velocity become zero so Vfy =0

0 = Vi Sin theta + gt

t = Vi Sintheta /g

Total time taken to land will be twice of that

On earth

Te= 2t

Te = 2Sinθ/g

on moon g is one-sixth of g(earth)

Tm = 2Sinθ/(g/6)

Tm = 6(2Sinθ/g)

Tm = 6Te

so total time taken by the projectile on moon will be six times the time taken on earth

From first equation X = Vix*T

we can see that X will also be 6 times on moon than earth

so projectile will cover 6 times distance on moon than on earth

4 0

3 years ago

Read 2 more answers

A person who weighs 150 pounds on Earth would weigh ____ pounds on the moon.

enot [183]

Answer:

25lb

Explanation:

You haven't changed (you are made up of the same atoms), but the force exerted on you is different. Physicists like to say that your mass hasn't changed, only your weight.

6 0

3 years ago

A long copper wire of radius 0.321 mm has a linear charge density of 0.100 μC/m. Find the electric field at a point 5.00 cm from

krek1111 [17]

Answer:

$E=35921.96N/C$

Explanation:

From the question we are told that:

Radius $r=0.321mm$

Charge Density $\mu=0.100$

Distance $d= 5.00 cm$

Generally the equation for electric field is mathematically given by

$E=\frac{mu}{2\pi E_0r}$

$E=\frac{0.100*10^{-6}}{2*3.142*8.86*10^{-12}*5*10^{-2}}$

$E=35921.96N/C$

4 0

3 years ago

When the green arrow and solid red light is illuminated, __________?

otez555 [7]

<span>When the green arrow and solid red light is illuminated, </span>means you turn in the direction of the arrow.

5 0

3 years ago

Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m

insens350 [35]

To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

$F_{mag}= BIL$

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: $F_{mag}$ is a direct adaptation of the vector relation $F=q \times V \times B$ </em>

From Newton's second law we know that the relation of Strength and weight is determined as

$F_g = mg$

Where,

m = Mass

g = Gravitational Acceleration

For there to be balance the two forces must be equal therefore

$F_{mag} = F_g$

$BIL = mg$

Our values are given as,

Diameter $(d) = 1.0mm = 1*10^{-3}m$

Radius $(r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m$

Magnetic Field $(B) = 5.0*10^{-5} T$

From the relationship of density another way of expressing mass would be

$\rho = \frac{m}{V} \rightarrow m = \rho V$

At the same time the volume ratio for a cylinder (the shape of the wire) would be

$V = \pi r^2 L \rightarrow L =Length, r= Radius$

Replacing this two expression at our first equation we have that:

$BIL = mg$

$BIL = ( \rho V)g$

$BIL = ( \rho \pi r^2 L)g$

Re-arrange to find I

$I = \frac{( \rho \pi r^2 L)g}{BL}$

$I = \frac{( \rho \pi r^2 )g}{B}$

We have for definition that the Density of copper is $8.9*10^3 Kg/m^3$ , gravity acceleration is $9.8m/s^2$ and the values of magnetic field (B) and the radius were previously given, then:

$I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}$

$I = 1370.05A$

The current is too high to be transported which would make the case not feasible.

8 0

3 years ago